Solveeit Logo
Login

Question

Question: If \(x = \frac{1 - t^{2}}{1 + t^{2}}\)and \(y = \frac{2t}{1 + t^{2}}\), then \(\frac{dy}{dx} =\)...

If x=1−t21+t2x = \frac{1 - t^{2}}{1 + t^{2}}and y=2t1+t2y = \frac{2t}{1 + t^{2}}, then dydx=\frac{dy}{dx} =

A

−yx\frac{- y}{x}

B

yx\frac{y}{x}

C

−xy\frac{- x}{y}

D

xy\frac{x}{y}

Answer

−xy\frac{- x}{y}

Explanation

Solution

x=1−t21+t2x = \frac{1 - t^{2}}{1 + t^{2}} and y=2t1+t2y = \frac{2t}{1 + t^{2}}

Put t=tan⁡θt = \tan\theta in both the equations, we get

x=1−tan⁡2θ1+tan⁡2θ=cos⁡2θx = \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} = \cos 2\theta and y=2tan⁡θ1+tan⁡2θ=sin⁡2θy = \frac{2\tan\theta}{1 + \tan^{2}\theta} = \sin 2\theta.

Differentiating both the equations, we get dxdθ=−2sin⁡2θ\frac{dx}{d\theta} = - 2\sin 2\theta and dydθ=2cos⁡2θ.\frac{dy}{d\theta} = 2\cos 2\theta.

Therefore dydx=−cos⁡2θsin⁡2θ=−xy\frac{dy}{dx} = - \frac{\cos 2\theta}{\sin 2\theta} = - \frac{x}{y}.